Problem: $ F = \left[\begin{array}{rrr}3 & 2 & 1 \\ -1 & 3 & 1\end{array}\right]$ $ B = \left[\begin{array}{rr}-2 & 5 \\ 0 & 3 \\ 0 & -2\end{array}\right]$ What is $ F B$ ?
Answer: Because $ F$ has dimensions $(2\times3)$ and $ B$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ F B = \left[\begin{array}{rrr}{3} & {2} & {1} \\ {-1} & {3} & {1}\end{array}\right] \left[\begin{array}{rr}{-2} & \color{#DF0030}{5} \\ {0} & \color{#DF0030}{3} \\ {0} & \color{#DF0030}{-2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ B$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ B$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ B$ , and so on. Add the products together. $ \left[\begin{array}{rr}{3}\cdot{-2}+{2}\cdot{0}+{1}\cdot{0} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ B$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{-2}+{2}\cdot{0}+{1}\cdot{0} & ? \\ {-1}\cdot{-2}+{3}\cdot{0}+{1}\cdot{0} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ B$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{-2}+{2}\cdot{0}+{1}\cdot{0} & {3}\cdot\color{#DF0030}{5}+{2}\cdot\color{#DF0030}{3}+{1}\cdot\color{#DF0030}{-2} \\ {-1}\cdot{-2}+{3}\cdot{0}+{1}\cdot{0} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{3}\cdot{-2}+{2}\cdot{0}+{1}\cdot{0} & {3}\cdot\color{#DF0030}{5}+{2}\cdot\color{#DF0030}{3}+{1}\cdot\color{#DF0030}{-2} \\ {-1}\cdot{-2}+{3}\cdot{0}+{1}\cdot{0} & {-1}\cdot\color{#DF0030}{5}+{3}\cdot\color{#DF0030}{3}+{1}\cdot\color{#DF0030}{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-6 & 19 \\ 2 & 2\end{array}\right] $